Suppose that X and Y are independent and \(Z = X + Y\). Find \(f_Z\) if
\[ f_x(x) = \begin{cases} \lambda^{-\lambda x} \text{ if } x>0\\ 0 \text{ otherwise}\\ \end{cases}\\ f_y(x) = \begin{cases} \mu e^{-\mu y} \text{ if } x>0\\ 0 \text{ otherwise} \end{cases} \]
\[ f_z(Z) = \int_{-\infty}^{\infty} f_x (z-y) f_y(y)dy\\ = \int_{0}^{\infty} f_x(z-y)f_y(y)dy = \int_{0}^{z} f_x(z-y)f_y(y) dy\\ = \int_{0}^{z} \lambda e^{-(z-y)}\mu e^{-\mu y} dy= \lambda \mu \int_{0}^{z} e^{-\lambda z + \lambda y} e^{-\mu y}dy\\ = \lambda \mu \int_{0}^{z} e^{-\lambda z} e^{\lambda y} e^{-\mu y}dy = \lambda \mu e^{-\lambda z}\int_{0}^{z}e^{y(\lambda -\mu)}dy\\ = \lambda \mu e^{-\lambda z} \int_{0}^{z} e^{-y(\mu - \lambda)} dy= \frac{\lambda \mu e^{-\lambda z}}{\mu - \lambda} [e^{-y(\mu - \lambda)}]_0^z\\ = \frac{\lambda \mu e^{-\lambda z}}{\mu - \lambda} [1- e^{-z(\mu - \lambda)}] = \frac{\lambda \mu e^{-\lambda z} - \lambda \mu e^{-\lambda z - z \mu + \lambda z}}{\mu - \lambda}\\ = \frac{\lambda \mu e^{-\lambda z} - \lambda \mu e^{-z \mu}}{\mu - \lambda} = \frac{\lambda \mu}{\mu - \lambda} (e^{-\lambda z}- e^{-z\mu})\\ f_z(x) = \begin{cases} \frac{\lambda \mu}{\mu - \lambda} (e^{-\lambda x}- e^{-x\mu}) \text{ when } x>0\\ 0 \text{ otherwise}\\ \end{cases}\\ \]
\[ f_x(x) = \begin{cases} \lambda^{-\lambda x} \text{ if } x>0\\ 0 \text{ otherwise}\\ \end{cases}\\ f_y(x) = \begin{cases} 1 \text{ if } 0<x<1\\ 0 \text{ otherwise} \end{cases} \]
When \(0<Z=X+Y<1\):
\[ f_z(Z) = \int_{-\infty}^{\infty} f_x(z-y) f_y(y) dy\\ = \int_{0}^{1} f_x(z-y) f_y(y) dy = \int_0^z \lambda e^{-\lambda(z-y)}dy \\ = \int_0^z \lambda e^{-\lambda z}e^{\lambda y}dy = \lambda e^{-\lambda z} \int_0^z e^{\lambda y}dy \\ = \frac{\lambda e^{-\lambda z}}{\lambda} [1-e^{\lambda y}]_0^z = e^{-\lambda z}(e^{\lambda z } -1) \\ = 1- e^{-\lambda z} \]
When \(Z>1\):
\[ f_z(Z) = \int_{-\infty}^{\infty} f_x(z-y)f_y(y)dy =\\ \int_0^1 f_x(z-y)f_y(y)dy = \int_0^1 \lambda e^{-\lambda(z-y)}\cdot 1 dy\\ = \int_0^1 \lambda e^{-\lambda z} e^{\lambda y}dy = \lambda e^{-\lambda z} \int_0^1 e^{\lambda y}dy\\ =\frac{\lambda e^{\lambda z}}{\lambda} [e^{\lambda y}]_0^1 = e^{-\lambda z}(e^{\lambda}-1) = e^{-\lambda z + \lambda} - e^{-\lambda z}\\ = e^{-\lambda (z-1)} - e^{-\lambda z} \]
In conclusion:
\[ f_z(x) = \begin{cases} 1 - e^{-\lambda x} \text{ when } 0 \leq x \leq 1\\ e^{-\lambda (x-1)} - e^{-\lambda x} \text{ when } x > 1\\ 0 \text{ otherwise} \end{cases} \]